Question

trapezoid defg is dilated according to the rule $d_{o, 4}(x, y)$ to form the image trapezoid defg, which is shown on the graph.
what are the coordinates of point g of the pre - image?
$(-2, 1)$
$(-32, 16)$
$(-4, 2)$
$(-12, 0)$

Explanation:

Step1: Recall Dilation Rule

The dilation rule is \( D_{O,4}(x,y) \), meaning the image coordinates \((x',y')\) are related to pre - image coordinates \((x,y)\) by \( x' = 4x \) and \( y' = 4y \), or \( x=\frac{x'}{4} \), \( y = \frac{y'}{4} \).

Step2: Find Coordinates of \( G' \)

From the graph, the coordinates of \( G' \) are \((-8,4)\).

Step3: Calculate Pre - image Coordinates of \( G \)

Using the dilation formula for pre - image (\(x,y\)) from image (\(x',y'\)): \( x=\frac{x'}{4} \), \( y=\frac{y'}{4} \).
For \( x'=-8 \), \( x=\frac{-8}{4}=-2 \).
For \( y' = 4 \), \( y=\frac{4}{4}=1 \).
So the coordinates of \( G \) (pre - image) are \((-2,1)\).

Answer:

\((-2,1)\)